Tasks for bitwise operations: And(&): 1510&1010 1011112&10001,2510&2410 1011111,&111011, OR(): 1510|1010o 101111,|10001,2510l2410 1011111,|1110112 Exclusive-OR (^): 1510^1010 101111,^10001, 2510^24101011111;^1110112 The bitwise shift operators(>2<<): 1510<<2 101111,>>3 2510<<3 1011111,>>4 The bitwise complement operators(~): ~(1510) 101111,) (2510) - (1011111,)

Tasks for bitwise operations: And(&): 1510&1010 1011112&10001,2510&2410 1011111,&111011, OR(): 1510|1010o 101111,|10001,2510l2410 1011111,|1110112 Exclusive-OR (^): 1510^1010 101111,^10001, 2510^24101011111;^1110112 The bitwise shift operators(>2<<): 1510<<2 101111,>>3 2510<<3 1011111,>>4 The bitwise complement operators(~): ~(1510) 101111,) (2510) - (1011111,)

Database System Concepts

7th Edition

ISBN:9780078022159

Author:Abraham Silberschatz Professor, Henry F. Korth, S. Sudarshan

Publisher:Abraham Silberschatz Professor, Henry F. Korth, S. Sudarshan

Chapter1: Introduction

Section: Chapter Questions

Problem 1PE

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Correct answer will be upvoted else Multiple Downvoted. Don't submit random answer. Computer science. Think about a diagram on k hubs, with numbers from b1 to bk composed on them. For each I from 1 to k: track down such j (1≤j≤k, j≠i), for which (bi⊕bj) is the littlest among all such j, where ⊕ indicates the activity of bitwise XOR (https://en.wikipedia.org/wiki/Bitwise_operation#XOR). Then, draw an undirected edge between vertices with numbers bi and bj in this chart. We say that the succession is acceptable if and provided that the subsequent diagram frames a tree (is associated and doesn't have any straightforward cycles). It is conceivable that for certain numbers bi and bj, you will attempt to add the edge between them twice. All things considered, you will add this edge just a single time. You can track down a model underneath (the image comparing to the main experiment). Succession (0,1,5,2,6) isn't great as we can't arrive at 1 from 5. Notwithstanding,…
Differentiate between • Symbolic addresses, Relocatable addresses and Absolute addresses • Dynamic loading and dynamic linking
The token bucket algorithm is a popular method for traffic shaping. A computer uses a token bucket with acapacity of 400 megabytes (MB) and a rate of 10 MB/sec. The computer continuously generates20 MB/sec for transmission from time 0. Assume the bucket’s water level is 200 MB at time 0.1a. How long will it take to send 1000 MB? Show the steps.1b. Draw a figure showing how the bucket’s water level evolves
implement greatestBitPos function Compute a mask marking the most significant 1 bit. you are only allowed to use the following eight operators: ! ~ & ^ | + << >> “Max ops” field gives the maximum number of operators you are allowed to use to implement each function /* * greatestBitPos - return a mask that marks the position of the* most significant 1 bit. If x == 0, return 0* Example: greatestBitPos(96) = 0x40* Legal ops: ! ~ & ^ | + << >>* Max ops: 70* Rating: 4 */int greatestBitPos(int x) {return 2;}
In computing, time scales usually store the current time as the number of seconds that have elapsed since the beginning of some “epoch”, which is an arbitrary date that determines the starting time. For example, the Network Time Protocol (NTP) uses an epoch of midnight on January 1st (GMT) 1900. The elapsed seconds since the epoch are stored in a 32-bit unsigned integer. In many embedded systems, it is more practical to just consider the current year. In this case, the format is often modified such that 0 corresponds to midnight on January 1stof the current year. If a clock that was initialized to 0 at midnight of 1 January 2020 currently has a count of 3665044 seconds, what day and time does it represent? (For example, your answer should be in the form June 6, 8:47:12 am) Show your work
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Suppose a computer system uses 16-bit addresses for both its virtual and physical addresses. In addition, assume each page (and frame) has size 256 bytes. How many bits are used for the page number? How many bits are used for the offset? 8 bits each. With this system, what’s the maximum number of pages that a process can have? 256 Suppose that each entry in the page table comprises 4 bytes (including the frame number, the valid bit, and miscellaneous “bookkeeping bits”). An OS uses an array to store the page table. What is the size of the page table? 1024 Bytes Furthermore, suppose the first 6 pages of a process map to frames 222 to 227 (as decimal numbers), and the last 5 pages of the process map to frames 1 to 5 (also decimal numbers). All other pages are invalid. Draw the page table, including the valid bit and the frame number. DONE Translate the following virtual addresses to physical addresses, and show how you obtain the answers. (Hint: You do not need to convert…
Kocher’s timing attacks on cryptosystems illustrate this [1084]. Kocher notes that the instructions executed by implementations of cryptosystems depend on the setting of bits in the key. For example, the algorithm in Figure 9–7 implements a fast modular exponentiation function. If a bit is 1, two multiplications occur; otherwise, one multiplication occurs. The extra multiplication takes extra time. Kocher determines bits of the confidential exponent by measuring computation time x:= 1; atmp := a; for 1 := 0 to k-1 do begin If Zi = 1 then x := (x * atmp) mod n; atmp := )atmp * atmp) mod n; end; result := x; Figure 9–7 A fast modular exponentiation routine. This routine computes x = az mod n. The bits of z are zk–1... z0. Kocher’s attack derives information about the computation from the characteristic of time. As a cryptographic key is confidential, this is a side channel attack. This is an example of a passive side channel attack, because results are derived only from observations. The…
Define the following variant of the Fibonacci sequence called Fib3 in C programming language : ● fib3(0) = 0 ● fib3(1) = 1 ● fib3(2) = 2 ● fib3(n) = fib3(n-1) + 2*fib3(n-2) + 3*fib3(n-3) for all n>=3 That is, the sequence is 0, 1, 2, 4, 11, 25, 59, 142, 335, 796, 1892, 4489... uint64_t fib3(unsigned int n);Your function needs to compute and return the correct value for all n<50 under 1 second.
Create a MIPS assembly program using MARS to do RSA encryption/decryption. Get a numeric message from the user and encrypt it. Also, get a numeric message from the user and decrypt it. An RSA public key consists of a pair of numbers (n, e), and the message m is encrypted to a ciphertext c by calculating c = me (mod n). The RSA private key consists of a pair of numbers (n, d), and a ciphertext c is decrypted by calculating m = cd (mod n). For this, use n=21733, e=257, d=1403. Note that since me and cd can get rather large (hundreds or thousands of digits for our examples), use the following pseudocode to handle encryption and decryption: doRSA( n, exp, msg ) cmsg = msg Initialize ciphertext, requires odd exponent mpow = msg message to the power 1, 2, 4, 8, 16, … tempexp = exp temporary exponent while ( tempexp > 1 ) mpow = ( mpow * mpow ) % n square the message tempexp = tempexp / 2 update the power if ( tempexp % 2 == 1 ) 1 bit in the binary representation of e? cmsg = ( cmsg…
In classful addressing, the IP address space is divided into 5 classes. Indicate the classes of each of the following addresses expressed in binary. Indicate how the class was identified. 3.1.1 00000001 00001011 00001011 11101111 3.1.2 11000001 10000011 00011011 11111111 3.1.3 10100111 11011011 10001011 01101111 3.1.4 11110011 10011011 11111011 00001111
EXAMPLE: Kocher’s timing attacks on cryptosystems illustrate this [1084]. Kocher notes that the instructions executed by implementations of cryptosystems depend on the setting of bits in the key. For example, the algorithm in Figure 9–7 implements a fast modular exponentiation function. If a bit is 1, two multiplications occur; otherwise, one multiplication occurs. The extra multiplication takes extra time. Kocher determines bits of the confidential exponent by measuring computation time x:= 1; atmp := a; for 1 := 0 to k-1 do begin If Zi = 1 then x := (x * atmp) mod n; atmp := )atmp * atmp) mod n; end; result := x; Figure 9–7 A fast modular exponentiation routine. This routine computes x = az mod n. The bits of z are zk–1... z0. Kocher’s attack derives information about the computation from the characteristic of time. As a cryptographic key is confidential, this is a side channel attack. This is an example of a passive side channel attack, because results are derived only from…